What is not moving the asymptote under a shift?

Adding 1 to e^x raises the asymptote to y = 1; leaving it at y = 0 is wrong.

State the domain and vertical asymptote of y = _3 x. [2 marks]

SEAB Original-style practice: State the domain of y = ln(x - 2) and the equation of its vertical asymptote.

The logarithm is defined only when its argument is positive: x - 2 > 0, so x > 2. That is the domain. As x → 2^+ the argument tends to zero and ln tends to -∞, so the vertical asymptote is x = 2. Markers reward the domain x > 2 and the vertical asymptote x = 2.

SingaporeAdditional MathematicsSyllabus dot point

What do the graphs of exponential and logarithmic functions look like, and how are they related?

Sketch the graphs of exponential and logarithmic functions, identify their key features, and recognise them as reflections of each other

A focused answer to the O-Level A-Maths outcome on exponential and logarithmic graphs. Their shapes, intercepts, asymptotes, and the inverse relationship that reflects one in the line y equals x.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to sketch $y = a^x$ and $y = \log_a x$ (and shifted or natural versions such as $e^x$ and $\ln x$ ), to mark their intercepts and asymptotes, and to recognise that the exponential and logarithmic functions are inverses, so each is the reflection of the other in the line $y = x$ . A correct sketch is often worth several marks and underpins later equation-solving.

The answer

The exponential graph

For $a > 1$ , the graph of $y = a^x$ :

passes through $(0, 1)$ , since $a^0 = 1$ ,
increases, getting steeper as $x$ grows,
has the $x$ -axis ( $y = 0$ ) as a horizontal asymptote, approached as $x \to -\infty$ ,
is always positive, so it never touches or crosses the $x$ -axis.

The logarithmic graph

For $a > 1$ , the graph of $y = \log_a x$ :

passes through $(1, 0)$ , since $\log_a 1 = 0$ ,
increases, but more and more slowly,
has the $y$ -axis ( $x = 0$ ) as a vertical asymptote, approached as $x \to 0^+$ ,
is defined only for $x > 0$ .

The inverse relationship

Because $\log_a x$ undoes $a^x$ , the two graphs are reflections of each other in the line $y = x$ . A point $(p, q)$ on the exponential corresponds to $(q, p)$ on the logarithm. This swaps the intercept $(0,1)$ into $(1,0)$ and the horizontal asymptote into a vertical one.

Transformations

Adding a constant shifts the curve vertically (and moves a horizontal asymptote); replacing $x$ by $x - h$ shifts it horizontally (and moves a vertical asymptote). Read the new asymptote from the shift.

Worked example

Sketch $y = \ln x$ and $y = e^x$ on the same axes and describe their relationship.

Step 1: Plot the exponential

$y = e^x$ passes through $(0, 1)$ , increases, and has asymptote $y = 0$ to the left.

Step 2: Plot the logarithm

$y = \ln x$ passes through $(1, 0)$ , increases slowly, and has asymptote $x = 0$ from the right; it is defined only for $x > 0$ .

Step 3: Add the line of symmetry

Draw $y = x$ . Each curve is the reflection of the other in this line, so $(0,1)$ on $e^x$ matches $(1,0)$ on $\ln x$ .

Step 4: State the relationship

The functions are inverses: $\ln(e^x) = x$ and $e^{\ln x} = x$ for $x > 0$ , which is why the graphs are mirror images in $y = x$ .

Examples in context

Example 1. Cooling curves. A cooling object's temperature follows a decaying exponential approaching room temperature, so its graph is a falling curve with a horizontal asymptote at the ambient value, the physical meaning of the asymptote.

Example 2. Reading a log scale. Plotting earthquake energy against magnitude uses a logarithmic axis, where the slow growth of $\log x$ compresses a huge range of energies into a readable scale, exactly the shape of the logarithmic graph.

Try this

Q1. State the $y$ -intercept and asymptote of $y = 2^x$ . [2 marks]

Cue. $y$ -intercept $(0, 1)$ ; horizontal asymptote $y = 0$ .

Q2. State the domain and vertical asymptote of $y = \log_3 x$ . [2 marks]

Cue. Domain $x > 0$ ; vertical asymptote $x = 0$ .

Q3. Describe how the graph of $y = e^x - 3$ differs from $y = e^x$ . [2 marks]

Cue. Shifted down $3$ units; asymptote moves to $y = -3$ and the $y$ -intercept becomes $(0, -2)$ .

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksSketch the graph of

y = e^{x} + 1

, stating the equation of the horizontal asymptote and the

y

-intercept.

Show worked answer →

Start from $y = e^x$ , which passes through $(0, 1)$ and has asymptote $y = 0$ . Adding $1$ shifts the curve up by one unit.

The $y$ -intercept is at $x = 0$ : $y = e^0 + 1 = 2$ , so $(0, 2)$ .

The horizontal asymptote moves up to $y = 1$ , approached as $x \to -\infty$ . The curve rises steeply for positive $x$ .

Markers reward an increasing curve, the asymptote $y = 1$ , and the $y$ -intercept $(0, 2)$ .

Original3 marksState the domain of

y = \ln(x - 2)

and the equation of its vertical asymptote.